Problem 125 - Introductory Chemistry Atoms First

Question

You have two solutions, one \(0.800 \mathrm{M}\) sodium. phosphate and the other \(0.800 \mathrm{M}\) lead(II) acetate. (a) Write a net ionic equation for the precipitation reaction that occurs when these solutions are combined. (b) If you pour \(100.0 \mathrm{~mL}\) of the sodium phosphate solution and \(50.0 \mathrm{~mL}\) of the lead(II) acetate solution into the same flask, what is the theoretical yield of the precipitate in grams? (c) What is the molar concentration of the excess reactant ion?

Step-by-Step Solution

Answer

(a) The net ionic equation for the precipitation reaction between sodium phosphate and lead(II) acetate is: 2PO₄³⁻ (aq) + 3Pb²⁺ (aq) → Pb₃(PO₄)₂ (s) (b) The theoretical yield of the precipitate (Pb₃(PO₄)₂) is 10.8 g. (c) The molar concentration of the excess reactant (PO₄³⁻) ions is 0.355 M.

Step 1: Write a balanced chemical equation

The balanced chemical equation for the reaction between sodium phosphate (Na₃PO₄) and lead(II) acetate [Pb(CH₃COO)₂] is: 2Na₃PO₄ (aq) + 3Pb(CH₃COO)₂ (aq) → Pb₃(PO₄)₂ (s) + 6Na(CH₃COO) (aq)

Step 2: Write the net ionic equation

For the net ionic equation - first, we will write the ions in their aqueous states, then we will eliminate the spectator ions, and finally, write the net ionic equation. Ions in aqueous solution: 6Na⁺ (aq) + 2PO₄³⁻ (aq) + 3Pb²⁺ (aq) + 6CH₃COO⁻ (aq) → Pb₃(PO₄)₂ (s) + 6Na⁺ (aq) + 6CH₃COO⁻ (aq) Eliminate spectator ions Na⁺ and CH₃COO⁻ and write the net ionic equation: 2PO₄³⁻ (aq) + 3Pb²⁺ (aq) → Pb₃(PO₄)₂ (s)

Step 3: Calculate moles of reactants

First, find the moles of Na₃PO₄ and Pb(CH₃COO)₂ using their volumes and concentrations: Moles of Na₃PO₄=n1 =Volume × Concentration n1 = 0.100 L × 0.800 mol/L n1 = 0.0800 mol Moles of Pb(CH₃COO)₂ = n2 = Volume × Concentration n2 = 0.050 L × 0.800 mol/L n2 = 0.0400 mol

Step 4: Determine the limiting reactant

In the balanced chemical reaction, 2 moles of Na₃PO₄ react with 3 moles of Pb(CH₃COO)₂ to yield Pb₃(PO₄)₂. We need to compare their mole to mole ratio with respect to the balanced chemical equation. Divide the number of moles by their respective coefficients for both reactants: For Na₃PO₄: 0.0800 mol / 2 = 0.0400 For Pb(CH₃COO)₂: 0.0400 mol / 3 = 0.0133 The reactant with the lowest value (0.0133) is the limiting reactant. In this case, Pb(CH₃COO)₂ is the limiting reactant.

Step 5: Calculate the theoretical yield of the precipitate

To find the theoretical yield of Pb₃(PO₄)₂ (in grams), use stoichiometry and molar mass: From the balanced equation, 3 moles of Pb(CH₃COO)₂ produce 1 mole of Pb₃(PO₄)₂. The moles of Pb₃(PO₄)₂ = (1/3) x moles of Pb(CH₃COO)₂ = (1/3) x 0.0400 mol = 0.0133 mol Next, convert moles to grams: Mass of Pb₃(PO₄)₂ = moles × molar mass Molar mass of Pb₃(PO₄)₂ = 3×(207.2) + 2×(94.97) = 811.1 g/mol Mass = 0.0133 mol × 811.1 g/mol Mass = 10.8 g (rounded) The theoretical yield of the precipitate (Pb₃(PO₄)₂) is 10.8 g.

Step 6: Calculate the molar concentration of the excess reactant ion

Since Pb(CH₃COO)₂ is the limiting reactant, Na₃PO₄ is the excess reactant. We need to find the concentration of PO₄³⁻ ions left after the reaction. From the balanced equation, 2 moles of Na₃PO₄ react with 3 moles of Pb(CH₃COO)₂. The moles of Na₃PO₄ that reacted = (2/3) x moles of Pb(CH₃COO)₂ = (2/3) x 0.0400 mol = 0.0267 mol The moles of Na₃PO₄ left after the reaction = initial moles - reacted moles = 0.0800 mol - 0.0267 mol = 0.0533 mol As the volume of the final solution is 100 mL + 50 mL = 150 mL, the concentration of PO₄³⁻ ions can be calculated as: Molar concentration of PO₄³⁻ = moles / volume = 0.0533 mol / 0.150 L = 0.355 M The molar concentration of the excess reactant (PO₄³⁻) ions is 0.355 M.

Key Concepts

Limiting Reactant

In a chemical reaction, the limiting reactant is the substance that determines the amount of product that can be formed. It is entirely consumed when the reaction goes to completion. Once the limiting reactant is used up, the reaction stops, and no more products can form. This concept is crucial because it allows chemists to predict the maximum yield of a reaction.
To identify the limiting reactant, compare the mole ratios of the reactants based on the balanced equation. For the reaction between sodium phosphate (Na₃PO₄) and lead(II) acetate [Pb(CH₃COO)₂], we need to use their stoichiometric coefficients.

• The balanced equation is: 2Na₃PO₄ + 3Pb(CH₃COO)₂ → Pb₃(PO₄)₂ + 6Na(CH₃COO).
• Ensure that the correct mole ratio is maintained as per the equation. Here, 2 moles of Na₃PO₄ react with 3 moles of Pb(CH₃COO)₂.
• Calculate the mole-to-coefficient ratio for both reactants and identify which one has a smaller value.
• In this case, Pb(CH₃COO)₂ with a ratio of 0.0133 is the limiting reactant.

Recognizing the limiting reactant helps in efficiently planning chemical reactions to avoid waste and maximize yield.

Theoretical Yield

The theoretical yield is the maximum amount of product that can be generated from a chemical reaction, assuming complete conversion of the limiting reactant into the desired product. Calculating the theoretical yield is vital for understanding the efficiency of a reaction.
To find the theoretical yield:

• Use the balanced chemical equation to determine the stoichiometry of the reaction.
• Multiply the moles of the limiting reactant by the stoichiometry factor from the balanced equation to find the moles of the product.
• Then convert these moles of product into grams using the molar mass of the product.

For the reaction at hand, 3 moles of Pb(CH₃COO)₂ yield 1 mole of Pb₃(PO₄)₂. With Pb(CH₃COO)₂ as the limiting reactant:

• 0.0400 moles of Pb(CH₃COO)₂ provide \(\frac{1}{3}\) times that in moles of Pb₃(PO₄)₂, which equates to 0.0133 moles.
• Using Pb₃(PO₄)₂'s molar mass of 811.1 g/mol, the mass of the precipitate is calculated to be 10.8 g.

This shows that if the reaction goes perfectly, you can expect 10.8 grams of lead phosphate (Pb₃(PO₄)₂) as your theoretical yield.

Molar Concentration

Molar concentration, often expressed in moles per liter (M), indicates the number of moles of a solute present in one liter of solution. This is a fundamental concept in chemistry, as it helps to understand how much of a substance is dissolved in a solution.
To calculate molar concentration after a reaction, especially to find the concentration of any excess reactant, consider:

• The initial amount of the reactant present before the reaction.
• The volume of the solution after the reaction takes place.

In the case of the initial problem, sodium phosphate (Na₃PO₄) is the excess reactant. After the reactants have reacted:

• The excess moles of Na₃PO₄ remaining are 0.0533 moles.
• Considering a final solution volume of 150.0 mL (0.150 L), use the formula: molar concentration = moles / volume.
• This results in a concentration of 0.355 M for PO₄³⁻ ions.

This calculation of molar concentration is essential in ensuring proper reactant usage and for the potential application of the solution in further processes or reactions.